Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using u32 = unsigned;
using pr = std::pair<int, int>;
using vec = std::vector<pr>;
using std::cerr;
using std::cin;
using std::cout;
int n;
u32 a[10000], b[10000];
vec v0, v1;
inline void emplace(u32* a, vec& vc, int x, int y, int warn = 1) {
  if (warn && x == y) puts("please do not xor self");
  a[x] ^= a[y], vc.emplace_back(x, y);
}
inline void swap(u32* a, vec& vc, int x, int y) {
  if (x != y) {
    emplace(a, vc, x, y);
    emplace(a, vc, y, x);
    emplace(a, vc, x, y);
  }
}
int p0[32], p1[32];
inline void reduce(u32* a, vec& vc, int* pos) {
  for (int i = 0; i < 32; ++i) {
    pos[i] = -1;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 31; j >= 0; --j) {
      if (!a[i]) break;
      if (a[i] >> j & 1) {
        if (pos[j] == -1) {
          pos[j] = i;
          break;
        } else
          emplace(a, vc, i, pos[j]);
      }
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
  }
  reduce(a, v0, p0), reduce(b, v1, p1);
  for (int i = 0; i < 32; ++i) {
    if (p1[i] != -1 && p0[i] == -1) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 31; i >= 0; --i) {
    if (p0[i] != -1) {
      if (p1[i] == -1)
        emplace(a, v0, p0[i], p0[i], 0), p0[i] = -1;
      else {
        u32 cur = a[p0[i]] ^ b[p1[i]];
        for (int j = i - 1; j >= 0; --j)
          if (cur >> j & 1) {
            if (p0[j] != -1) {
              emplace(a, v0, p0[i], p0[j]);
              cur ^= a[p0[j]];
            } else {
              cout << -1 << '\n';
              return 0;
            }
          }
        int id = std::find(p0, p0 + 32, p1[i]) - p0;
        swap(a, v0, p0[i], p1[i]);
        if (id != 32) p0[id] = p0[i];
        p0[i] = -1;
      }
    }
  }
  v0.insert(v0.end(), v1.rbegin(), v1.rend());
  for (int i = 0; i < n; ++i) {
  }
  cout << v0.size() << '\n';
  for (auto i : v0) cout << i.first + 1 << ' ' << i.second + 1 << '\n';
}